Proof for $A,B \in M_n(\mathbb{F})$ that if $[A,B]=tA$ for $0\neq t\in\mathbb{F}$, then $A^n=0$

Question

Statement. Suppose we have a square matrices $A,B$ of order $n$ over a field $\mathbb{F}$ of characteristics $0$ or $p>n$. If $[A,B]=AB-BA=tA$ for some nonzero $t\in\mathbb{F}$, then $A^n=0$. The problem is to prove this statement.

Where I've got so far. There is an idea to take trace of both sides of equation: $$ \mathrm{tr}\,(AB-BA)=\mathrm{tr}\,(tA) $$ $$ \mathrm{tr}\,(AB)-\mathrm{tr}\,(BA)=t\times\mathrm{tr}\,(A) $$ since $\mathrm{tr}\,(AB) = \mathrm{tr}\,(BA)$ (by properties of the trace) we have $$ t\times A = \mathrm{tr}\,(A) $$ and since $t\neq 0$, we have $\mathrm{tr}\,(A) = 0$. That's all I could get form it. What sould I do next?

Answer 1

Hints.

1. Prove (by mathematical induction or otherwise) that $A^kB-BA^k=ktA^k$ for $k=1,2,\ldots$
2. The trace of a product of matrices is invariant under cyclical permutation of the multiplicands.
3. See the question "$\DeclareMathOperator{\tr}{tr}\tr(A)=\tr(A^{2})= \ldots = \tr(A^{n})=0$ implies $A$ is nilpotent". Several answers to this question are applicable when the underlying field has characteristic 0 or characteristic $p>n$.

Answer 2

As I mentioned in the comments, we need to impose $t\neq0$ in order for the statement to hold. As Najib, says, this statement can be proved using the previous statement you asked about.

We have$$t\cdot\mathrm{tr}(A)=\mathrm{tr}(tA)=\mathrm{tr}(AB-BA)=0,$$which yields$$\mathrm{tr}(A)=0.$$Likewise, for any $k$, we have $tA^k=A^kB-A^{k-1}BA.$ Using an argument similar to the one above, we deduce that for any $k$, $\mathrm{tr}(A^k)=0$. Hence, $A$ is nilpotent.